# `cos(theta)cot(theta)/(1 - sin(theta)) - 1 = csc(theta)` Verfiy the identity.

mathace | Certified Educator

Verify the identity `cos(theta)cot(theta)/(1-sin(theta))-1=csc(theta)`

`(cos(theta)cot(theta))/(1-sin(theta))-1=csc(theta)`

Rewrite `cot(theta)` as a quotient.

`[[cos(theta)/1]*[cos(theta)/sin(theta)]]/(1-sin(theta))-1=csc(theta)`

`[(cos^2(theta))/sin(theta)]/[(1-sin(theta)]]-[1]=csc(theta)`

`[(cos^2(theta))/sin(theta)]*1/(1-sin(theta))-1=csc(theta)`

Use the pythagorean identity `sin^2(theta)+cos^2(theta)=1` to substitute in for the `cos^2(theta).`

`[1-sin^2(theta)]/sin(theta)*[1/(1-sin(theta))]-1=csc(theta)`

Factor the term `1-sin^2(theta).`

`[(1+sin(theta)(1-sin(theta))]/sin(theta)]*[1/(1-sin(theta))]-1=csc(theta)`

Cancel the `(1-sin(theta)).`

`[(1+sin(theta))/sin(theta)]*[1]-1=csc(theta)`

Rewrite the first term as two separate fractions.

`1/sin(theta)+sin(theta)/sin(theta)-1=csc(theta)`

`csc(theta)+1-1=csc(theta)`

`csc(theta)=csc(theta)`

balajia | Student

`(cos(theta).cot(theta))/(1-sin(theta)) -1`

`= (cos^2(theta)(1+sin(theta)))/(sin(theta)(1-sin^2(theta)))-1`

`=(cos^2(theta)(1+sin(theta)))/(sin(theta)(cos^2(theta)))-1`

`=(1+sin(theta))/sin(theta)-1`

`=csc(theta)+1-1=csc(theta)`